Worked solution · Exam P
This Exam P sample reference tests Exponential Distribution. Each exponential operating time has variance 10 squared. Independence makes the total variance 100+100=200, choice E.
Independent solution
Setup
An exponential lifetime has variance equal to the square of its mean. Each operating time therefore has variance 100.
Model
The two operating times are independent, so the variance of their sum is the sum of their variances.
Compute
Adding the two equal contributions gives total variance 200.
Answer
The total operating time has variance 200, selecting choice E.
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30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.
Error diagnosis
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Original practice · fully worked
A data request passes through two independent service stages. Their times are exponential with means 4 seconds and 7 seconds. Find the variance of total service time.
Exponential variance equals mean squared. Adding the independent stage variances gives 16+49=65.
Setup
The stage variances are the squares of the exponential means, giving 16 and 49 seconds squared.
Model
Independence removes the covariance term, so the total service-time variance is the sum of the two stage variances.
Compute
The resulting variance is 65 seconds squared.
Answer
The total service time has variance 65 seconds squared, selecting choice D.
Keep practicing
These references share syllabus concepts or sit nearby in the official sequence. All 387 Exam P univariate random variables practice problems.
Drill the method: Exam P Exponential Distribution problems · Exam P Variance problems
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